The speed of a train is reduced from 80km/h to 40km/h after the application of the brake. (i)how much further would the train travel before coming to rest (ii)assuming the acceleration is kept constant,how long will it take to bring the train to rest after the application of the brakes?


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Question Number 19828 by NECC last updated on 16/Aug/17

$T h e s p e e d o f a t r a i n i s r e d u c e d$ $f r o m 80 k m / h t o 40 k m / h a f t e r$ $t h e a p p l i c a t i o n o f t h e b r a k e .$ $(i) h o w m u c h f u r t h e r w o u l d t h e$ $t r a i n t r a v e l b e f o r e c o m i n g t o r e s t$ $(i i) a s s u m i n g t h e a c c e l e r a t i o n i s$ $k e p t c o n s t a n t, h o w l o n g w i l l i t$ $t a k e t o b r i n g t h e t r a i n t o r e s t$ $a f t e r t h e a p p l i c a t i o n o f t h e b r a k e s ?$
Commented by NECC last updated on 16/Aug/17

$s o s o r r y . . . . T h e s p e e d o f a c a r w a s$ $r e d u c e d f r o m 80 k m / h t o 40 k m / h$ $f o r 500 m . . . . . . .$ $t h e o t h e r t e x t s a r e g i v e n$
	Answered by Tinkutara last updated on 16/Aug/17

	$v^{2} - u^{2} = 2 a s$ $a = \frac{40^{2} - 80^{2}}{2 \times 0.5} = - 4800 km / h^{2}$ $(i) s = \frac{v^{2} - u^{2}}{2 a} = \frac{0 - 1600}{- 2 \times 4800} = \frac{1}{6} km$ $Means, \frac{1}{6} km more after travelling$ $500 m already .$ $(i i) t = \frac{v - u}{a} = \frac{0 - 80}{- 4800} = \frac{1}{60} h = 1 \min$
	Commented by NECC last updated on 16/Aug/17

	$t h a n k s b o s s . . .$ $p l e a s e d o y o u k n o w a n y s i t e o r$ $p d f f i l e t h a t r e a l l y e x p l a i n s t h e$ $e q u a t i o n o f m o t i o n$
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